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Write an algorithm which computes the number of trailing zeros in n factorial.

Have you met this question in a real interview?? Yes Example

11! = 39916800, so the out should be 2

Challenge?

O(log N) time

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LeetCode上的原題，請(qǐng)參見(jiàn)我之前的博客Factorial Trailing Zeroes。

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解法一：

class Solution {public:// param n : description of n// return: description of return long long trailingZeros(long long n) {long long res = 0;while (n > 0) {res += n / 5;n /= 5;}return res;} };

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解法二：

class Solution {public:// param n : description of n// return: description of return long long trailingZeros(long long n) {return n == 0 ? 0 : n / 5 + trailingZeros(n / 5);} };

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轉(zhuǎn)載于:https://www.cnblogs.com/grandyang/p/6201067.html

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